Let $f(x)=\sqrt[3]{x^2-2x+8}$. Find $f'(2)$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $2$ (Choice C) C $\dfrac{8}{3}$ (Choice D) D $\dfrac{1}{6}$
Explanation: Let's start by finding the expression for $f'(x)$. Then, we can evaluate it at $x=2$. $f$ is a radical function, but its argument isn't simply $x$. Therefore, it's a composite radical function. In other words, suppose $u(x)=x^2-2x+8$, then $f(x)=\sqrt[3]{u(x)}$. $f'(x)$ can be found using the following identity: $\dfrac{d}{dx}\left[\sqrt[3]{u(x)}\right]=\dfrac{1}{3}[u(x)]^{^{-\scriptsize\dfrac{2}{3}}}u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\sqrt[3]{x^2-2x+8} \\\\ &=\dfrac{d}{dx}\sqrt[3]{u(x)}&&\gray{\text{Let }u(x)=x^2-2x+8} \\\\ &=\dfrac{1}{3}[u(x)]^{^{-\scriptsize\dfrac{2}{3}}}u'(x) \\\\ &=\dfrac{1}{3}[x^2-2x+8]^{^{-\scriptsize\dfrac{2}{3}}}(2x-2)&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ Now let's evaluate $f'( 2)$. $\begin{aligned} f'( 2)&=\dfrac{1}{3}\Bigl(( 2)^2-2( 2)+8\Bigr)^{^{-\scriptsize\dfrac{2}{3}}}\cdot\Bigl(2( 2)-2\Bigr) \\\\ &=\dfrac{1}{3}\cdot 8^{^{-\scriptsize\dfrac{2}{3}}}\cdot2 \\\\ &=\dfrac{1}{3}\cdot \dfrac{1}{4}\cdot 2 \\\\ &=\dfrac{1}{6} \end{aligned}$ In conclusion, $f'(2)=\dfrac{1}{6}$.